This worksheet provides answers and explanations for common kinematics of free fall problems. Understanding free fall is crucial for mastering introductory physics. Free fall, in the simplest terms, describes the motion of an object solely under the influence of gravity, neglecting air resistance. This means the object accelerates downwards at a constant rate, approximately 9.8 m/s² (often denoted as 'g') near the Earth's surface.
Note: These answers assume a constant acceleration due to gravity (g = 9.8 m/s²) and neglect air resistance. Results may vary slightly depending on the value of 'g' used.
Key Equations for Free Fall Kinematics
Before diving into the answers, let's review the fundamental kinematic equations that apply to free fall:
- v = u + at: Final velocity (v) equals initial velocity (u) plus acceleration (a) multiplied by time (t).
- s = ut + (1/2)at²: Displacement (s) equals initial velocity multiplied by time plus half the acceleration multiplied by the square of time.
- v² = u² + 2as: The square of the final velocity equals the square of the initial velocity plus twice the acceleration multiplied by the displacement.
Where:
- v = final velocity (m/s)
- u = initial velocity (m/s)
- a = acceleration due to gravity (g = -9.8 m/s² downwards, +9.8 m/s² upwards)
- t = time (s)
- s = displacement (m)
Example Problems and Solutions
To illustrate the application of these equations, let's work through some common free fall problems. Remember to always define your positive direction (usually upwards).
Problem 1: An object is dropped from a height of 100 meters. Calculate the time it takes to hit the ground.
Solution:
Here, u = 0 m/s (dropped, not thrown), s = -100 m (displacement is negative because it's downwards), and a = -9.8 m/s². We'll use the equation: s = ut + (1/2)at².
-100 = 0*t + (1/2)(-9.8)t²
Solving for t, we get t ≈ 4.52 seconds.
Problem 2: A ball is thrown vertically upwards with an initial velocity of 20 m/s. Calculate its maximum height.
Solution:
At the maximum height, the final velocity (v) will be 0 m/s. We'll use the equation: v² = u² + 2as.
0² = 20² + 2(-9.8)s
Solving for s, we get s ≈ 20.41 meters.
Problem 3: A stone is thrown downwards from a cliff with an initial velocity of 5 m/s. It hits the ground after 3 seconds. Calculate the height of the cliff.
Solution:
Here, u = -5 m/s (negative because it's thrown downwards), t = 3 s, and a = -9.8 m/s². We use the equation: s = ut + (1/2)at².
s = (-5)(3) + (1/2)(-9.8)(3)²
Solving for s, we get s ≈ -52.1 meters. The height of the cliff is approximately 52.1 meters. The negative sign indicates the displacement is downwards.
Problem 4: An object is thrown upwards and returns to the same height after 6 seconds. What was its initial velocity?
Solution: The time to reach the maximum height is half the total time (3 seconds). At the maximum height, v = 0. Using v = u + at:
0 = u + (-9.8)(3)
Solving for u, we get u ≈ 29.4 m/s.
Further Practice and Resources
These examples provide a solid foundation. Further practice with varied problems is key to mastering free fall kinematics. You can find additional practice problems and worked examples in introductory physics textbooks and online resources. Remember to always clearly define your coordinate system and pay close attention to the signs of your variables (positive and negative directions). Consistent practice will build confidence and proficiency in solving these types of problems.
